Solution Manual for Chemical, Biochemical, And Engineering Thermodynamics – Stanley Sandler

Access full Solution Manual only here Solutions to Chemical and Engineering Thermodynamics, 5 th ed

Chapter 4

http://www.book4me.xyz/solution-manual-for-chemical-biochemical-and-engineering-thermodynamics-sandler/ 

4 4.1 

System = Ball (1) + Water (2)

(a)

Energy balance:  M1U1 f + M2U 2f − M1U1i − M2U 2i = 0 







⇒  M1CV,1cT1 f − T1i h + M2 CV,2 cT2f − T 2i h = 0 ; also T1 f − T 2f  . Thus  f 

T  =

 M1CV,1T1i + M2 CV,2 T 2i  M1CV,1 + M2 C V,2

=

5 × 103 × 0.5 × 75 + 12 × 103 × 4.2 × 5 5 × 103 × 0.5 + 12 × 103 × 4.2

= 8.31° C [Note: Since only ∆T  s are involved, ° C  were used instead of K)]. dT  T  (b) For solids and liquids we have (eqn. 4.4-6). 4.4-6). That ∆S = M C P =  MC P ln 2  for the case in T  T 1

z 

which C P  is a constant. Thus Ball: ∆S  = 5 × 10 1 03 g × 0.5

J g ⋅ K 

31 + 273.15 U J R 8.31  = −53161 . V K  T 75 + 273.15 W

× lnS

= −53161 .  s Water: ∆S  = 12 × 103 g × 4.2

J g ⋅ K 

31 + 273.15 U J R 8.31  = +59622 . V K  T 5 + 273.15 W

× lnS

and ∆S (Ball + Water )   = 596.22 − 531.61

4.2

J

= 64.61

J

K  K   Note that the system system Ball + Water is isolated. isolated. Therefore J ∆S = S gen = 6461 . K  Energy balance on the combined system of casting and the oil bath

c

h

c

h

 M cCV,c T  f  − Tci + Mo CV,o T  f  − T oi = 0  since there is a common final temperature.

20 kg × 0.5

kJ

kJ

dT  f − 450iK + 150 kg × 2.6 kg ⋅ K dT f  − 450iK = 0 kg ⋅ K 

This has the solution T f  = 60oC = 313.15 K Since the final temperature is known, the change in entropy of this system can be calculated 273.15 + 60 I  F 273.15 + 60I  . kJ from ∆S  =  = 20 × 0.5 × lnF  H 273.15 + 450K  + 150 × 2.6 × lnH 273.15 + 50K  = 4135 K 

4.3

Closed system energy and entropy balances dU 

 dV 



dt 



dt 



;

dS 

Q 

+ S gen ; dt  T   dS  Thus, in general Q = T  − TS gen and dt  = Q + Ws − P

=





Solutions to Chemical and Engineering Thermodynamics, 5 th ed

W s =

dU 

−Q+ P



 dV 



dt 

dt 

=

Chapter 4

dU 

 dS   dV  − T  + TSgen + P dt  dt  dt  

dU 

 dS   dV  − T  +P dt  dt  dt  dU  dV  (a) System at constant U  &  & V  ⇒ = 0  and =0 dt  dt 

Reversible work: WsRev = WsRev c S gen = 0h = 





WscSgen = 



dS 

(b) System at constant S  &  & P ⇒

dt 

Rev Rev

h = WS

0



= −T

dP

= 0  and

 dS dt

= 0⇒ P

dt 

 dV  dt 

=

d  dt 

( PV )

so that

c

h

dU 

Ws S gen = 0 = W Srev  = 





dt 

+

d  dt 

( PV ) =

d  dt 

(U + PV ) =

dH  dt 

4.4

700 bar, 600oC

10 bar, T  =  = ?

Steady-state balance equations dM  = 0 =  M1 + M 2 dt  

dU 



0 dV 0 =  M1 H1 + M2 H 2 = 0 =  M1H1 + M2 H 2 + Q 0 + Ws − P 

















dt

dt 

or  H1 = H 2 



Drawing a line of constant enthalpy on Mollier Diagram we find, at P  = 10  bar, T  ≅  ≅ 308° C At 700 bar and 600° C

At 10 bar and 308° C

3

V = 0.003973 m kg

V  ≈ 0.2618 m3 kg

 H = 3063  kJ kJ kg

H  ≈ 3063 kJ kg

S = 5.522 kJ kg K

S  = 7.145 kJ kg K  













Also dS 

⇒ Sgen = M1c S 2 − S1h   or 



0





dt 

Q T

= 0 =  M1S1 + M2 S 2 +









S gen



+ S gen = 0 



 M 1 

4.5

= S2 − S 1 = 7.145 − 5.522 = 1.623 



1

System

2

Ws

Energy balance ∆U = cU 2 f − U2i h + cU1f − U1i h  = Q 

adiabatic

+ W S  −

z 

constant

PdV volume

kJ kg ⋅ K 

Solutions to Chemical and Engineering Thermodynamics, 5 th ed

c

h

c

h

Chapter 4

c

h c

h

Ws = MC p T2 f − T2i + MCp T1 f − T1i = MCp T2 f − T2i + T1 f − T1i  

 but T1 f = T2f = T  f  ⇒

W s

f

i

i

= 2T − T1 − T 2

 MC P

Entropy balance adiabatic

∆S = c S2 f − S2i h + cS1f − S1i h  =

Q 0 dt +Sgen T

z 



0 for maximum work

cS2 f

− S2i h + cS1f − S1i h = 0 = MCP l n

R T2 f T 1 f  U V=0; T T2i T 1i W

or lnS

T 2 f  T 2i

+  MC P ln

T 1 f  T 1i

T1 f T2f = T1i T2i ;   but T1 f = T2f = T  f 

2

⇒ cT  f h = cT1i T 2i h  or T  f = T1i T 2i  and W s  MC P

4.6

= 2T f − T1i − T2i = 2 T1i T2i − T1i − T2i  

1 bar

10 bar

290 K

575 K

(a) Entropy change per mole of gas ∆ S = C P ln

T 2 T 1

−  R ln

P2

eqn. (4.4-3)

P1

575 J 10 J ln − 8314 . ln = 09118 . mol K  290 mol K  1 mol K  (b) System = contents of turbine (steady-state system) dN  Mass balance = 0 =  N1 + N 2 ⇒ − N 2 = N1 = N  dt  dV 0 dU  0 Energy balance + Ws − P = 0 =  N1 H 1 + N2 H 2 + Q dt dt  Ws = N a H 2 − H1 f = NCP a T2 − T1 f  Thus ∆S  =  = 29.3

J











W s 

W  =

 N  















= CP aT2 − T 1f = 29.3

J mol K 

× (575 − 290)K 

J

= 83505 .

mol (c) In Illustration Illustration 4.5-1, 4.5-1, W  =  = 78348 . J mol   because of irreversiblitities ( ∆S  ≠  ≠ 0) , more work is done on the gas here. What happens to this this additional energy energy input? It appears as an increase of the internal energy (temperature) of the gas.

4.7

Heat loss from metal block dU  dT  = C P =Q dt  dt  

Solutions to Chemical and Engineering Thermodynamics, 5 th ed

−W  = 

C P

T − T 2 T 

RQ = heat out of metal S−Q = heat into heat engine T 

Q(−1) 



dT  aT − T 2 f dt 

T 

T 2

t 

z

z F H 1 − T  I K dT 



0

LF  NH 

T 2

= −W ⇒ − Wdt = CP   

T 1

−W = CP aT2 − T1f − CP ⋅ T2 l  n −W = CPT 2 MG1 −

T 2 T 1

L N

= CP MaT2 − T1f − T2 l n

T 2 O T 1 PQ

T 1 I  T  O  −  − ln 2 P J  T 2 K  T 1 Q

T 2

Q=

Chapter 4

F 

T 1 I 

z C dT = C aT − T f = C T H G  1 − T  J K  P

P

2

1

P 2

2

T 1

Alternate way to solve the problem

T 2

System is the metal block + heat engine (closed) dU  dT  E.B.: = C P = Q + W  dt  dt  

S.B.:

dS  dt 

Q





=

T 2

Q = T 2

+Sgen

dS 



0 for maximum work

;

dU 

= T 2

dS 

+ W  ; dU = CPdT  ; dS  = 

C P

dt  dt  dt  T  dU  dS  C P T  F  I  W  = − T 2 = CP dT − T2  dT = C P 1 − 2 dT  H  dt  dt  T  T  K 

dT 



T 2

T 2

1

1

F  T  I  F  T  I  W = z Wdt = z CP  1 − 2 dT = C P z 1 − 2 dT  H  T  K  H  T  K  T  T  T  T  O LF  T  I  − W = CP aT2 − T1 f − T2 CP  ln 2 = CP T 2 MG1 − 1 J   − ln 2 P T 1 NH  T 2 K  T 1 Q 

4.8

This problem is not well posed since since we do not know exactly exactly what is happening. There are several several  possibilities: (1) Water contact is is very short so neither neither stream changes T   very much. In this case we have the the Carnot efficiency  = η  =

 −W  Q

=

Thigh − T low T high

=

22 27 + 273

=

22 300

= 0.07 0733 = 7.33 33%

(2) Both warm surface water water (27° (27°C) and cold deep water (5° (5 °C) enter work producing device, and they leave at a common temperature.

T H T O T L

Solutions to Chemical and Engineering Thermodynamics, 5 th ed

Chapter 4

http://www.book4me.xyz/solution-manual-for-chemical-biochemical-and-engineering-thermodynamics-sandler/  dM 

M.B.:

= 0 =  M H + M L + M0 ⇒ M0 = −b M H + M L g 

dt  dU 

E.B.:











= 0 =  M H HH + M L HL + M0 H0 + W  = 0 



dt 











b g =  M H c H0 − HH h + M L c H0 − H L h =  M H CP aT0 − TH f + M LCP aT0 − T L f

W = − M H HH − M L HL + M H + M L H0   































Q = 0 =  M H SH + M L SL + M0 S 0 + T dt 

dS 

S.B.:

0











b

+Sgen



g

0

 M H SH + M L SL − M H + ML S 0 = 0 



c





h



c





h

 M H SH − S0 + M L SL − S0 = 0 ⇒ M H C P ln 







F T H I  M F T L I M  H G T 0 J K  H G T 0 J K  



H

 M H

T0  = TH







T H T 0

+  M LC P ln 

T L T 0

=0

L

= 1 or TH M H TL M L = T 0 M H + M L 

b M H + ML g 





ML

T L









b M H + M L g 



From this can calculate T 0 . Then

a

f

a

f

W = M H CP T0 − TH + M L CP T0 − TL   





This can be used for any flow rate ratio. (3) Suppose very large amount of surface water water is contacted with a small small amount of deep water, i.e., i.e.,  M H >> M L . Then T0 ~ T H 



a

f

a

f

a

f

W = M H CP TH − TH + ML CP TH − TL ~ ML CP TH − TL   







(4) Suppose very large amount of deep water water is contacted with a small small amount of surface water, i.e., i.e.,  M H